1. Technical Field
The disclosure relates to a circuit for protecting an electronic device against malfunctions caused by undervoltage conditions, and more particularly to undervoltage protection circuit for LED lamp.
2. Description of the Related Art
For years, people have used traditional incandescent or fluorescence lighting apparatus in order to address their interior lighting concerns. However, such lighting apparatus presents a number of drawbacks. For example, the popular halogen apparatus present the following drawbacks, such as relatively high power consumption, inefficiency of light dispersion due to the placement of its metal shield in the line sight of the halogen bulb, and its limited effectiveness in preventing glare from the halogen bulb.
Recently, a number of LED lamp have been designed to replace the halogen apparatus, as well as other traditional incandescent or fluorescence lighting apparatuses, which are utilized in some commercial lighting, such as exhibition cabinet, horizontal freezer etc. As well known, the LED lamp must be powered by an appropriate supply, such as constant flow source or constant voltage source. It is important when operating the LED lamp that the supply which powers the device be carefully monitored. In particular, the supply voltage must be maintained within a tolerance range necessary to ensure proper operation of the LED lamp. If the supply voltage deviates outside the tolerance range, then the device may malfunction or, worse yet, may be destroyed. For example, DC/DC converter is generally used in the constant voltage source. However, the DC/DC converter has no 100% duty ratio. In result, there is voltage difference between the input voltage and the output voltage. For example, it is assumed that output voltage of the LED lamp is 18V, and then, the input voltage of the DC/DC converter need to be 24V. As a result, 6V of voltage difference may be formed therebetween. When the DC/DC converter works and the load is full, input voltage will rise from 0V until to 24V. However, while the input voltage rises into 18V, the DC/DC converter begins to work, and now, there is a little voltage difference therebetween and the DC/DC converter has a big duty ratio. As the output voltage rises, the input current of the DC/DC converter will increase and the input current of the supply will increase, which not conform to safety requirement.